Problem 2: A2Okg boxis pushed 5 meters up a 350 incline ramp bya 20 N force. The force is parallel to the incline: The kinetic coefficient of friction between the r...

Block Slides Down an Incline In Fig. $10-48$, a block is moved down an incline a distance of $5.0 \mathrm{~m}$ from point $A$ to point $B$ by a force $\vec{F}$ that is parallel to the incline and has magnitude $2.0 \mathrm{~N}$. The magnitude of the frictional force acting on the block is $10 \mathrm{~N}$. If the kinetic energy of the block increases by $35 \mathrm{~J}$ between $A$ and $B$, how much work is done on the block by the gravitational force as the block moves from $A$ to $B$ ?

So we can first use Equation 8 31 and we can obtain that the change in thermal energy would be equal to the kinetic frictional force multiplied by the distance. We know that this is gonna be equal to 10 Newtons multiplied by 5.0 meters. So this is equaling 50 Jules. And then we're going to use a quite equation 78 In order to obtain that the work is equaling the force times the distance. This would be equal to 2.0 Newton's multiplied by 5.0 meters. This would be equal to 10 Jules and then we can say that again Equation 8 31 Ah would give us that the work is equaling the change in kinetic energy, plus the change in thermal energy. Sorry, plus the changing potential energy plus the plus the change in thermal energy and this would be equal to rather reconsidered. 10 jewels would be equal to 35 jewels, plus the change in thermal energy plus 50 50 jewels. And we can then say that the change in thermal energy sorry the change in potential energy would be equal to negative 75 jewels. So therefore we can say that the work done by gravity would be equal to the negative change and potential energy. This is equaling, of course. Positive. 75 jewels. This would be the work done by gravity. So our final answer, the work done by gravity is 75 jewels. That is the end of the solution. Thank you for watching.

Again this question. A block of mass two kg exposed down an inclined plane. Okay, So there is an inclination. Okay. And mass of two kg. M as opposed to two kg is given. Okay. And the inclination is 37 degrees. So three to is given 37 degree. And it is posed down with the force of continued. And so it will act like this. So F is given 20 new town. Okay. And it is founded. The block moves on the inclination with an exploration of 10 m per second squared 10 m par second square. Okay. And it is started from the rest. Okay, so no problem. First of all, we can find out the distance covered by this block as that is you people as half it is square. Okay, so initially it is coming from the rest, so you will be zero. Also, as will be only one by 2 80 square and one by two exploration is given 10 m per second square and it is in the 1st 2nd. Okay, So it is one. So it will be as it was 25 m. Okay? So in one second it will travel or the distance will be five m. Okay. And now party of this question. Okay. We have to find out the work done by the Applied force. Okay? So what we've done that is forced into distance and only by the Applied Force Applied forces F. And the distant government that is as supplied force F. That is given 20 newton. And the distance covered that is five m. So it will be hundreds. Rule, that is the answer of part A of the discussion. Okay. Now part B. By the weight of the block in the 1st 2nd. Okay? So by the weight. Okay. So we will go downward. Okay. That is the world done will be M G. H. Okay. And heritage will be if this distance that is five m and this is angle 37 degree. So this will be the edge and edge will be five sign 37. Okay so MG. That is mars is to Angie's 10 and edge will be five sign 37°.. Okay so it will be two and 10 2020 and five multiplied by. It will be 100 multiple. I've assigned 37 that is three by five. Okay so it will be 20 and 20 multiple laboratory that is 60 jewel. Okay and the last part work done by the frictional force. Okay and again work done is fourth multiplied by distance. But here the force that is frictional force. Okay so in this part okay the friction. Okay it will be MG. Scientist to Okay because MG will be here downward. So MG scientist I will be the frictional force. Okay? So MG scientist are multiplied by Yes, that is the distance. So it will be five so M. G. Again to multiply it by 10. Multiplied by scientists. That is three by five. Multiplied by five. So when we saw it will be again 60 years old. Okay? So this is the answer of part A. This is part of the and this is part C. Thank you.

Here, somebody has propelled a block down an incline with a speed of 1.6 m per second. And the block slides a distance of 1.1 oh meters before coming to a stop. And the question is, what is the coefficient of kinetic friction on that surface? So to answer that, we are going to need both Kinnah Matics and Newton's second law. So a reminder of what those are. And then we'll get into the kingdom attics, which usually is a little bit easier to do. But Newton's second law is the sum of the forces equals mass times acceleration. And there are three equations usually that you use to work with constant acceleration. I am going to concentrate on the one equation that does not involve time because I see that no time is given in the problem. So I'm going to use the final squared minus v. Initial squared is equal to twice the acceleration times the distance. And we're going to pick a cord, that system with why pointing up and X pointing down along the incline some and with that as our situation we have the final squared zero and of course the initial is 1.60 m per second, which we will square is to times the unknown acceleration times the distance of sliding 1.1 oh meters is down the incline. So what this will provide for us is that the acceleration is negative and specifically it is in the minus X. Direction and kind of working that out. I it winds up to be 1.60 squared with a negative in front divided by twice 1.1 oh. And let me sub label that with an X. So I realize that that is an exhilaration along the incline, hence force dubbed the extraction. And that comes out to be minus 1.16 m per second squared. And why I did that first is knowing that Newton's second law is going to involve the acceleration. And typically what you want to do is break this into two equations, one for the X. Direction and one for the Y direction. So decompose your vectors and you really can't use Newton's second law until you have a force diagram and then you can start working with the components. So let's create a force diagram. So I have my mass, which I know and I'm going to create a weight downwards, that's always good place to start. And I'm going to show that that weight ex at an angle. So I'm going to kind of show my axes in there. I have an X. Axis and a Y axis. Let's make that a different color. And my weight actually makes an angle of 30 degrees to my Y. Axis. That is it becomes zero if I am flat on a horizontal, completely horizontal um oriented surface. So that's the easy one. The other easy one to draw is the normal force is straight along the y. Axis. And finally there is a force of fiction which is going to act opposite the motion and it is sliding friction. So we will call that F. Sub K. And now I can decompose things into components and set up my Newton's second law. So I usually make a column for my some of Fx and another column for my some of fy. And then I take each force separately and work with it sometimes starting with the easy ones. Um So the force of friction for example we know is completely along the X. Direction. It's up the incline which is in the negative direction. And knowing what I do about friction, It is U. K. Times the force of the normal, Let's be consistent. We call that normal voice. Good idea and now I'm happy I see my unknown there um and it has no Y. Component so whoops. Yeah no why component good? And my normal force is again the easy an easy one to draw out. It is no zero, no X. Component and all lying along the Y. Direction. And finally the weight is the one that has components but I also know a lot about it. We know the mass, we know 9.8 m per second squared. Um And the X. Component goes with the sign of 30 degrees and the why component is downwards and goes with the co sign of 30 degrees and I'll give myself a little bit of space to work. It's always a good idea um to actually work out numbers because we do need numbers. Okay, so the X component working it out with, the sign of 30 degrees is 12.25 newtons. And the Y component working out with the co sign is minus yeah, 21.2. And then at the bottom we know everything has to add up to zero so we can write an equation in our bottom a table element where we've added everything together and set it equal to zero. I'm sorry though, it's equal to zero in the Y direction but it's equal to M A X in the extraction. Yeah, so that will produce two equations. Um And hopefully only two unknowns and I can identify my unknowns fairly readily that they are going to be the coefficient of friction and the normal force and the algebra is not going to be too terrible, we hope. Um so what we have is minus U sub k times normal force plus 12.25 buttons is equal to M A X. And we'll put in a X. Because we do know that quantity on the other side, we have fn minus 21.2. Newtons equals to zero. And as promised, the algebra is not too difficult to do um equation number two. There can be solved for fN fairly readily. 21.2 newtons, we can use it in our first equation and solve for the coefficient of friction. Yeah. Mhm minus UK times 21.2 newtons. I'll leave off the units just so things don't look so complicated and all that has to equal the mass times the acceleration which we found above. Mhm Okay. And I'll show one. An intermediate step are actual, Yeah, just show one intermediate step. And so we have a minus U. K. Yeah. That doesn't even dane. Another step. We can turn all our negatives into a positive and we get a coefficient of friction fairly readily from that arithmetic.

Mhm. This problem covers the concept of the Newton's 2nd law of motion. And to solve that, we have drawn three border diagram of the system on the left hand side. Uh Since the object is moving with constant speed so far, I'm one week a diet. The submission of all forces along the direction of motion. It is the longest direction equals zero. Uh um One G minus T equals zero. Other danger equals. And one time she let's say this is the question one. Now for M. two, you can read the submission of all forces along the plane. Sorry, Along the plane is X. and the proper divert. The plane is five. The submission of the oil forces along The vertical direction perpendicular to the surface equals zero. And also the submission of all forces along the plane is C. Now by equating the net force along the vertical direction at zero we get and two equals Mdg cost theater. And by creating the net force along the X Direction is zero. That tension minus MG. Scientific to monastic kinetic friction force equals C. Okay, now from these two equations we can buy it. And the question one from all these equations you can right M two G minus and one G. Into science theater less milky cost cedar equals and one less. Uh Sorry, made a mistake. Okay, so from this equation we can right and one G minus I. M two G into milk it. I'm scores theater less scientific data equals C Of the Moss and one equals I am going to milky horse to bless scientific, I have stood the value So the mass and one equals and do is 10 Kg. And do milkers? 0.2 Cause of 30° plus sign off 100 degrees Or the mass and one equals 6.73 kg